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anastassius [24]

3 years ago

5

An arrow in flight has an initial velocity of 65 meters per second, and 10 seconds later, it has a velocity of 35 meters per sec

ond. Which is the acceleration of the arrow?

1 answer:

-BARSIC- [3]3 years ago

7 0

Acceleration = (final velocity - initial velocity) / time
= (35-65)/10
= -3 m/s2

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What type of eclipse is shown? How do you know?

Sonja [21]

That is a lunar eclipse. At night, when the Earth is between the Sun and the moon, the moon would appear to be red. Just for future reference, a solar eclipse is when the Moon is between the Sun and Earth. Speaking of which, check out the solar eclipse this August!

8 0

3 years ago

Read 2 more answers

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$

5 0

3 years ago

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful

balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

$|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T$

the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

4 0

2 years ago

Calculat the acceleration of a person at latitude<br> 40degreesowing to the rotation of the earth.

beks73 [17]

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

so

acceleration of person = g - ω²R ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × $10^{-5}$ rad/s and R is earth radius that is 6.37 × $10^{6}$ m

so

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × $10^{-5}$ )² × 6.37 × $10^{6}$

acceleration of person = 9.77 m/s²

6 0

3 years ago

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

Akimi4 [234]

Answer:

a) The distance of spectator A to the player is 79.2 m

b) The distance of spectator B to the player is 43.9 m

c) The distance between the two spectators is 90.6 m

Explanation:

a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:

x = v * t

where:

x = position of the spectators

v = speed of sound

t = time

Then, the position for spectator A relative to the player is:

x = 343 m/s * 0.231 s = 79.2 m

b)For spectator B:

x = 343 m/s * 0.128 s

x = 43.9 m

The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.

c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:

(Distance AB)² = A² + B²

(Distance AB)² = (79.2 m)² + (43.9 m)²

Distance AB = 90. 6 m

6 0

3 years ago